The only possible option here without an even difference already is option 2, and for option 2 'U' would be even (due to the fact that C and A will either both be even or both be odd).

Therefore without even taking the fourth number into consideration, we can tell that the product of all the differences is always even. (because we only need one even in the differences, as anything times an even is even!)

Using this information we can now determine that for any given sequence of numbers above 2, the product of their differences will always be even!! :-)

The product of the first three number differences is even therefore the product of the differences of four+ numbers wil also be even, because anything times an even will always be even. So no matter how many numbers there are (over 3) the product of the differences will always be even.

To work this out we have to work out the minimum number of even differences there can be:
If the minimum is 2 then the product of the differences has to be a multiple of 4,
if the minimum is 3 then the product of the differences has to be a multiple of 8,
if the minimum is 4 then the product of the differences has to be a multiple of 16,...

What I did to work this out was:
I made the gaps between the five numbers all odd. (In a sequence a, b, c, d, e. a-b=odd, b-c=odd, etc.)
This then...

... caused a-c, b-d and c-e to be even (odd+odd=even)
a-e would also be even (odd+odd+odd+odd=even)
This would also make a-d odd and b-e odd.

Therefore the minimum number of even numbers in the differences would be 4.

So for a sequence of 5 numbers the product of all their differences would be at least a multiple of 16.